So I'm currently studying universal algebraic geometry on a basis of a few arXiv papers, and I'm trying to gain a bit of intuition concerning coordinate algebras, so I tried computing one example that was not given and that seems quite natural.
Right now I don't have the link of the paper at hand but if you need it, I can provide it a bit later.
For an algbra $A$ and an algebraic set $Y$ over $A$, $\Gamma_A(Y)$ is the coordinate algebra of $Y$, that is the term algebra with $n$ generators (where $Y\subset A^n$) with the relations in $\operatorname{Rad}_A(Y)$ (those are the identities satisfied by the points in $Y$).
For a set $S$ of identities in $n$ variables, $V_A(S)$ is the set of points in $A^n$ satisfying those identities.
I'm trying to compute $\Gamma_{\mathbb{R}}(S^1)$ where $S^1$ is the circle, with the language being that of rings.
The term algebra with the identities holding in any ring generated by $x,y$ is $\mathbb{Z}[x,y]$. Adding the relations of $S^1$, we get that it's a quotient of $\mathbb{Z}[x,y]/(x^2 + y^2 -1)$ since $S^1 = V_{\mathbb{R}}(x^2 + y^2 =1 )$. I'd like to know if $\Gamma_{\mathbb{R}}(S^1) \simeq \mathbb{Z}[x,y]/(x^2+y^2 -1)$.
I'm not quite sure about this so I'd like your help there. What I can say is that if one can show that $B=\mathbb{Z}[x,y]/(x^2 + y^2 -1) \simeq \Gamma_{\mathbb{R}}(Y)$ for some $Y\subset \mathbb{R}^2$ then $Y\subset S^1$ and so$Y=S^1$ because $S^1$ does have a coordinate algebra.
But I don't know how to prove this.
So essentially my questions are :
- Is my intuition the right one, i.e. is $\mathbb{Z}[x,y]/(x^2 + y^2 -1)$ actually the coordinate algebra of $S^1$ ? If so, how can I prove it ? If not, what is ?
-What's the intuition behind coordinate algebras, what do they represent, why do we want to know them ?
I'm adding this remark : my first question comes down to : if for all $t$, $P(\cos(t), \sin(t))=0$, is it then true that $x^2+ y^2 -1 \mid P(x,y)$ ?
It turns out that my remark led me to the answer, so I'm answering my own question as I now know how to.
If $P\in\Bbb{Z}[X,Y]$ vanishes on $S^1$, then, doing a euclidean division of $P$ by $Y^2 + X^2 -1 $ in $\Bbb{Z}[X][Y]$ (possible since $Y^2 + X^2 -1 $ is unitary in $Y$) we can restrict our attention to polynomials of the form $R(X)Y + Q(X)$ which vanish on $S^1$. It is then easy to see that such a polynomial is $0$, which shows that $X^2 + Y^2 -1$ divides $P$ in $\mathbb{Z}[X,Y]$.
As I've noted, this implies $\Gamma_{\mathbb{R}}(S^1) \simeq \mathbb{Z}[x,y]/(x^2 + y^2 -1)$